A hard puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13911 Accepted Submission(s): 4872
Problem Description
lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.
this puzzle describes that: gave a and b,how to know the a^b's the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise.
Input
There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30)
Output
For each test case, you should output the a^b's last digit number.
Sample Input
7 66
8 800
Sample Output
9
6
Author
eddy
题意:求a^b结果的个位数是多少
#include <iostream>
using namespace std;
/*
分析:当a的个位是0 1 2 3 4 5 6 7 8 9时,只会出现一下循环,例如个位是2 那以后个位只会在2 4 8 6之间变动
0
1
2 4 8 6
3 9 7 1
4 6
5
6
7 9 3 1
8 4 2 6
9 1
*/
int cns[10][4] = {{0},{1}, {6, 2, 4, 8},{1, 3, 9, 7},{6, 4}, {5}, {6}, {1, 7, 9, 3}, {6, 8, 4, 2}, {1, 9}};
int main()
{
int a, b, digit, cnt;
while(cin>>a>>b)
{
digit = a % 10;
switch(digit)
{
case 0:
case 1:
case 5:
case 6:
cout<<digit<<endl;
continue;
break;
case 2:
case 3:
case 7:
case 8:
cnt = b % 4;
break;
case 4:
case 9:
cnt = b % 2;
break;
}
cout<<cns[digit][cnt]<<endl;
}
return 0;
}
分享到:
相关推荐
HDU的1250,主要是利用高精度加法,但是代码有点繁琐,效率不是很高
杭电ACMhdu1163
HDU1059的代码
hdu1001解题报告
hdu 1574 passed sorce
HDU的一题........HDU DP动态规
hdu2101AC代码
hdu acm 教案 搜索入门 hdu acm 教案 搜索入门
搜索 dfs 解题代码 hdu1241
hdu 5007 Post Robot 字符串枚举。 暴力一下就可以了。
hdu_2102_passed_sorce
hdu acm 教案 动态规划(1) hdu acm 教案 动态规划(1)
现在,给你两个正的小数A和B,你的任务是代表大明计算出A+B的值。 Input 本题目包含多组测试数据,请处理到文件结束。 每一组测试数据在一行里面包含两个长度不大于400的正小数A和B。 Output 请在一行里面...
hdu1290 解题报告 献给杭电五十周年校庆的礼物 (切西瓜问题,即平面分割空间)
ACM HDU题目分类,我自己总结的大概只有十来个吧
hdu 1166线段树代码
HDU最全ac代码
自己做的HDU ACM已经AC的题目
hdu动态规划算法集锦